Calculating the dimension of a cotangent space of a curve












2












$begingroup$


Let $A$ be the local ring
$$A=left(frac{mathbb{C}[x,y]}{(y^2-x^3-x^2)}right)_{(x,y)}$$
$mathfrak{m}=(x,y)A$ be the maximal ideal in $A$. I'm asked to calculate the dimension of the $mathbb{C}$-vector space $mathfrak{m}/mathfrak{m}^2$ which is the dimension of the cotangent space at $(0,0)$. I do it as follows.



$$
begin{aligned}
mathfrak{m}/mathfrak{m}^2&=left(frac{(x,y)}{(y^2-x^3-x^2)}right)_{(x,y)}/left(frac{(x,y)^2}{(y^2-x^3-x^2)}right)_{(x,y)}\
&=left(frac{(x,y)_{(x,y)}}{(y^2-x^3-x^2)_{(x,y)}}right)/left(frac{(x,y)^2_{(x,y)}}{(y^2-x^3-x^2)_{(x,y)}}right)\
&=frac{(x,y)_{(x,y)}}{(x,y)^2_{(x,y)}}=left(frac{(x,y)}{(x,y)^2}right)_{(x,y)}simeqmathbb{C}^2
end{aligned}
$$
This seemingly correct argument gives the desired result 2, but I realize that I didn't use any information about $(y^2-x^3-x^2)$, and it will be the same at any other point on $y^2-x^3-x^2$ which is impossible. I think there must be some problem about my argument, but I cannot find it out. Where am I wrong? What is the correct way to do it? Thanks in advance.






algebraic-geometry commutative-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    You might find my answer here useful.
    $endgroup$
    – André 3000
    Dec 13 '18 at 5:03








  • 1




    $begingroup$
    @André 3000 It helps, thanks.
    $endgroup$
    – Y.Guo
    Dec 13 '18 at 5:26










  • $begingroup$
    Actually, I just realized that this question deals with the same ring as in your question.
    $endgroup$
    – André 3000
    Dec 13 '18 at 5:28










  • $begingroup$
    @Y.Guo Please explain why "it will be the same at any other point on $y^2-x^3-x^2$ which is impossible".
    $endgroup$
    – user26857
    Dec 13 '18 at 22:06
















2












$begingroup$


Let $A$ be the local ring
$$A=left(frac{mathbb{C}[x,y]}{(y^2-x^3-x^2)}right)_{(x,y)}$$
$mathfrak{m}=(x,y)A$ be the maximal ideal in $A$. I'm asked to calculate the dimension of the $mathbb{C}$-vector space $mathfrak{m}/mathfrak{m}^2$ which is the dimension of the cotangent space at $(0,0)$. I do it as follows.



$$
begin{aligned}
mathfrak{m}/mathfrak{m}^2&=left(frac{(x,y)}{(y^2-x^3-x^2)}right)_{(x,y)}/left(frac{(x,y)^2}{(y^2-x^3-x^2)}right)_{(x,y)}\
&=left(frac{(x,y)_{(x,y)}}{(y^2-x^3-x^2)_{(x,y)}}right)/left(frac{(x,y)^2_{(x,y)}}{(y^2-x^3-x^2)_{(x,y)}}right)\
&=frac{(x,y)_{(x,y)}}{(x,y)^2_{(x,y)}}=left(frac{(x,y)}{(x,y)^2}right)_{(x,y)}simeqmathbb{C}^2
end{aligned}
$$
This seemingly correct argument gives the desired result 2, but I realize that I didn't use any information about $(y^2-x^3-x^2)$, and it will be the same at any other point on $y^2-x^3-x^2$ which is impossible. I think there must be some problem about my argument, but I cannot find it out. Where am I wrong? What is the correct way to do it? Thanks in advance.






algebraic-geometry commutative-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    You might find my answer here useful.
    $endgroup$
    – André 3000
    Dec 13 '18 at 5:03








  • 1




    $begingroup$
    @André 3000 It helps, thanks.
    $endgroup$
    – Y.Guo
    Dec 13 '18 at 5:26










  • $begingroup$
    Actually, I just realized that this question deals with the same ring as in your question.
    $endgroup$
    – André 3000
    Dec 13 '18 at 5:28










  • $begingroup$
    @Y.Guo Please explain why "it will be the same at any other point on $y^2-x^3-x^2$ which is impossible".
    $endgroup$
    – user26857
    Dec 13 '18 at 22:06














2












2








2


0



$begingroup$


Let $A$ be the local ring
$$A=left(frac{mathbb{C}[x,y]}{(y^2-x^3-x^2)}right)_{(x,y)}$$
$mathfrak{m}=(x,y)A$ be the maximal ideal in $A$. I'm asked to calculate the dimension of the $mathbb{C}$-vector space $mathfrak{m}/mathfrak{m}^2$ which is the dimension of the cotangent space at $(0,0)$. I do it as follows.



$$
begin{aligned}
mathfrak{m}/mathfrak{m}^2&=left(frac{(x,y)}{(y^2-x^3-x^2)}right)_{(x,y)}/left(frac{(x,y)^2}{(y^2-x^3-x^2)}right)_{(x,y)}\
&=left(frac{(x,y)_{(x,y)}}{(y^2-x^3-x^2)_{(x,y)}}right)/left(frac{(x,y)^2_{(x,y)}}{(y^2-x^3-x^2)_{(x,y)}}right)\
&=frac{(x,y)_{(x,y)}}{(x,y)^2_{(x,y)}}=left(frac{(x,y)}{(x,y)^2}right)_{(x,y)}simeqmathbb{C}^2
end{aligned}
$$
This seemingly correct argument gives the desired result 2, but I realize that I didn't use any information about $(y^2-x^3-x^2)$, and it will be the same at any other point on $y^2-x^3-x^2$ which is impossible. I think there must be some problem about my argument, but I cannot find it out. Where am I wrong? What is the correct way to do it? Thanks in advance.






algebraic-geometry commutative-algebra







share|cite|improve this question









$endgroup$




Let $A$ be the local ring
$$A=left(frac{mathbb{C}[x,y]}{(y^2-x^3-x^2)}right)_{(x,y)}$$
$mathfrak{m}=(x,y)A$ be the maximal ideal in $A$. I'm asked to calculate the dimension of the $mathbb{C}$-vector space $mathfrak{m}/mathfrak{m}^2$ which is the dimension of the cotangent space at $(0,0)$. I do it as follows.



$$
begin{aligned}
mathfrak{m}/mathfrak{m}^2&=left(frac{(x,y)}{(y^2-x^3-x^2)}right)_{(x,y)}/left(frac{(x,y)^2}{(y^2-x^3-x^2)}right)_{(x,y)}\
&=left(frac{(x,y)_{(x,y)}}{(y^2-x^3-x^2)_{(x,y)}}right)/left(frac{(x,y)^2_{(x,y)}}{(y^2-x^3-x^2)_{(x,y)}}right)\
&=frac{(x,y)_{(x,y)}}{(x,y)^2_{(x,y)}}=left(frac{(x,y)}{(x,y)^2}right)_{(x,y)}simeqmathbb{C}^2
end{aligned}
$$
This seemingly correct argument gives the desired result 2, but I realize that I didn't use any information about $(y^2-x^3-x^2)$, and it will be the same at any other point on $y^2-x^3-x^2$ which is impossible. I think there must be some problem about my argument, but I cannot find it out. Where am I wrong? What is the correct way to do it? Thanks in advance.






algebraic-geometry commutative-algebra




algebraic-geometry commutative-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 13 '18 at 3:15









Y.GuoY.Guo

1619




1619












  • $begingroup$
    You might find my answer here useful.
    $endgroup$
    – André 3000
    Dec 13 '18 at 5:03








  • 1




    $begingroup$
    @André 3000 It helps, thanks.
    $endgroup$
    – Y.Guo
    Dec 13 '18 at 5:26










  • $begingroup$
    Actually, I just realized that this question deals with the same ring as in your question.
    $endgroup$
    – André 3000
    Dec 13 '18 at 5:28










  • $begingroup$
    @Y.Guo Please explain why "it will be the same at any other point on $y^2-x^3-x^2$ which is impossible".
    $endgroup$
    – user26857
    Dec 13 '18 at 22:06


















  • $begingroup$
    You might find my answer here useful.
    $endgroup$
    – André 3000
    Dec 13 '18 at 5:03








  • 1




    $begingroup$
    @André 3000 It helps, thanks.
    $endgroup$
    – Y.Guo
    Dec 13 '18 at 5:26










  • $begingroup$
    Actually, I just realized that this question deals with the same ring as in your question.
    $endgroup$
    – André 3000
    Dec 13 '18 at 5:28










  • $begingroup$
    @Y.Guo Please explain why "it will be the same at any other point on $y^2-x^3-x^2$ which is impossible".
    $endgroup$
    – user26857
    Dec 13 '18 at 22:06
















$begingroup$
You might find my answer here useful.
$endgroup$
– André 3000
Dec 13 '18 at 5:03






$begingroup$
You might find my answer here useful.
$endgroup$
– André 3000
Dec 13 '18 at 5:03






1




1




$begingroup$
@André 3000 It helps, thanks.
$endgroup$
– Y.Guo
Dec 13 '18 at 5:26




$begingroup$
@André 3000 It helps, thanks.
$endgroup$
– Y.Guo
Dec 13 '18 at 5:26












$begingroup$
Actually, I just realized that this question deals with the same ring as in your question.
$endgroup$
– André 3000
Dec 13 '18 at 5:28




$begingroup$
Actually, I just realized that this question deals with the same ring as in your question.
$endgroup$
– André 3000
Dec 13 '18 at 5:28












$begingroup$
@Y.Guo Please explain why "it will be the same at any other point on $y^2-x^3-x^2$ which is impossible".
$endgroup$
– user26857
Dec 13 '18 at 22:06




$begingroup$
@Y.Guo Please explain why "it will be the same at any other point on $y^2-x^3-x^2$ which is impossible".
$endgroup$
– user26857
Dec 13 '18 at 22:06










1 Answer
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$begingroup$

I’ll be really sloppy here; you can clean up after me.



Consider the nonsingular point $P=(-1,0)$, so that you want to localize at $(x+1,y)$. Your generators seem to be $x+1$ and $y$, but since $x$ is a unit in the localization, you could generate the maximal ideal just as well by $(x+1)x^2$ and $y$. In other words, by $y^2$ and $y$, thus by $y$ alone.






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    1 Answer
    1






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    active

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    active

    oldest

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    4












    $begingroup$

    I’ll be really sloppy here; you can clean up after me.



    Consider the nonsingular point $P=(-1,0)$, so that you want to localize at $(x+1,y)$. Your generators seem to be $x+1$ and $y$, but since $x$ is a unit in the localization, you could generate the maximal ideal just as well by $(x+1)x^2$ and $y$. In other words, by $y^2$ and $y$, thus by $y$ alone.






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      I’ll be really sloppy here; you can clean up after me.



      Consider the nonsingular point $P=(-1,0)$, so that you want to localize at $(x+1,y)$. Your generators seem to be $x+1$ and $y$, but since $x$ is a unit in the localization, you could generate the maximal ideal just as well by $(x+1)x^2$ and $y$. In other words, by $y^2$ and $y$, thus by $y$ alone.






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        I’ll be really sloppy here; you can clean up after me.



        Consider the nonsingular point $P=(-1,0)$, so that you want to localize at $(x+1,y)$. Your generators seem to be $x+1$ and $y$, but since $x$ is a unit in the localization, you could generate the maximal ideal just as well by $(x+1)x^2$ and $y$. In other words, by $y^2$ and $y$, thus by $y$ alone.






        share|cite|improve this answer









        $endgroup$



        I’ll be really sloppy here; you can clean up after me.



        Consider the nonsingular point $P=(-1,0)$, so that you want to localize at $(x+1,y)$. Your generators seem to be $x+1$ and $y$, but since $x$ is a unit in the localization, you could generate the maximal ideal just as well by $(x+1)x^2$ and $y$. In other words, by $y^2$ and $y$, thus by $y$ alone.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 13 '18 at 3:39









        LubinLubin

        44.7k44586




        44.7k44586






























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